rezbipul sum 1

[chestnut.cc]

x/|x| < x . Which of the following must be true about x ?
A)x>1
B)x>-1
C)|x|<1
D)|x|=1
E)|x|^2>1
Ans:
two interpretation of the question.
1. If the given f(x) is true , then what is the value of x ?
2. For what value of x, a given function f(x) is true ?

There is a subtle difference between two.
I took the first interpretation, others the second one.

It is given as true x/|x| < x , it is asked what must be true about x which is equivalent to finding out the range of x where this holds, this gives only two range -1<x<0 and x>1,
Now from the given alternatives I am looking for where these both conditions hold good.

A. x>1 , one of the condition is true, other one not
B. x > -1, both hold
C. |x| = 1 , not in above range
D. |x| < 1 , gives -1<x<1 , not good ,
E. |x|^2 > 1 ,( &#166;x&#166;^2 = x^2 that means x^2>1 => x>1 or x<-1) x<-1, x>1 , so one holds other not

For x>1 the inequality will always hold good. So the obvious solution to this problem should be A but there are some values less than 1 which also satisfies the inequality as question asks for "Which of the following must be true about X?" so whatever is the case x will always be greater than -1 as there is no value less than -1 which can satisfy the equation.


OR,

x/x < x or x/-x < x
1< x or -1 < x
x is greater than 1 or x is greater than -1
=> Combining both: x is always greater than -1. Hence the answer.

OR,

X / |X| < X
Multiply both sides by |X| (We can multiple by |X|, since |X| is always greater than zero).

X < X|X|
=> X - X|X| < 0
=> X (1 - |X| ) < 0

Now, there are two cases, Case 1 OR case 2

Case 1:
X < 0 and 1 - |X| > 0
=> X < 0 and 1 + X > 0 (Notice that |X| = - X since we know that X < 0)
=> X < 0 and X > -1
=> X lies between -1 and 0

Case 2:
X > 0 and 1 - |X| < 0
=> X > 0 and 1 - X < 0 (Notice that |X| = X since we know that X > 0)
=> X > 0 and X > 1
=> X > 1

So now we have,
Case 1 OR Case 2
=> X lies between -1 and 0 OR X > 1

Going through the sets that rezbipul uploaded... came across this one... imo its a bit overworked... heres my soln...

x/|x| < x... x!=0 |x|!=0... Multiplying by |x|>0... x<x|x|... x^2< x^2*|x|^2... |x|^2 = x^2 hence x^2 < x^4... x^2 >0 hence we can divide... 1 <x^2... now if x^2 >1... then x >1 or x <-1... the only ans is E... :s

[e.cartman]

x<x|x|... x^2< x^2*|x|^2

you are squaring both sides of inequality, but you dont know that both sides are positive. how did you assume that?

[chestnut.cc]

x/|x| < x... this inequality will only hold if x > -1 or x >1... x >1 means x >-1...
x*|x| = -x^2 or x*|x| = x^2... either way that inequality still holds cos x >-1... so you can square without compunction... the ans then becomes x > -1 though... so thank you for pointing that out... :s

[grexam]

Hi
I always get stuck with inequality problems ......can anybody please post a collection of problems on inequalities concept.........I find them really difficult to solve.......any suggestions???

[chestnut.cc]

dl the rezbipul problem set... it has problems and explanations...

[grexam]

@chestnut

can u please be more precise what exactly is rezbipul????...........I tried but couldnot find them

thanks a lot

[chestnut.cc]

the thread about a few places below this one...